1.

50 ml of 0.05 M HNO_(3)is added to 50 mlof 0.024 M KOH. Calculate the pH of theresultantsolution

Answer»

Solution :Numberofmolesof ` HNO_(3) = 0.05 XX 50 xx 10^(-3)`
`=2.5 xx 10^(-3)`
Number of MOLES of KOH = ` 0.025 xx 50 xx 10^(-3)`
` 1.25 xx 10^(-3)`
Number of moles of ` HNO_(3)`after mixing
2.5 xx 10^(-3) - 1.5 xx 10^(-3)`
`1.25 xx 10^(-3)`
CONCETRATION of ` HNO_(3)`
` = ( " Number of molesof ` NHO_(3)) /( " Volume of littre") `
After mixing , totalvolume =` 100 ml = 100 xx 10^(-3) L `
` [ H^(+)] = ( 1.25 xx10^(-3) " moles")/( 100 xx 10^(-3) L)`
` 1.25 xx 10^(-2) " moles " L^(-1)`
PH= `log [ H^(+)]`
` pH= - log( 1.25 xx 10^(-2)) = 2- 0.0969`
` = 1.9031 `


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